Question

Find the values of $x$ which satisfy the equation: $4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)$

• A. $x=1,4,\frac{1}{8}$
• B. $x=2,6,\frac{1}{8}$
• C. $x=3,4,\frac{1}{8}$
• D. None of these

Answer 1

The correct option is A $x=1,4,\frac{1}{8}$

$4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)$
$\Rightarrow \frac{4{\log }_2\sqrt{x}}{{\log }_2\left(\frac{x}{2}\right)}+\frac{2{\log }_2\left(x^2\right)}{{\log }_2\left(4x\right)}=\frac{3{\log }_2\left(x^3\right)}{{\log }_2\left(2x\right)}[\because {\log }_{b}a=\frac{\log a}{\log b}]$
$\Rightarrow$ $\frac{4\times \frac{1}{2}{\log }_2\left(x\right)}{{\log }_{2}x-1}+\frac{4{\log }_2\left(x\right)}{2+{\log }_2\left(x\right)}=\frac{9{\log }_2\left(x\right)}{1+{\log }_2\left(x\right)}[\because \log a^m=m\log a\text{\&}\log (ab)=\log a+\log b\text{\&}{\log }_{a}a=1]$
Let ${\log }_{2}x=t$. The given equation reduces to
$\frac{2t}{t-1}+\frac{4t}{t+2}=\frac{9t}{t+1}$
or $t=0$ or $\frac{2}{t-1}+\frac{4}{t+2}=\frac{9}{t+1}$
or $\frac{2t+4+4t-4}{(t-1)(t+2)}=\frac{9}{t+1}$
or $t^2+t-6=0$
or $(t+3)(t-2)=0$
or $t=0,2$ 0r $-3$
$\Rightarrow$ $x=1,4,\frac{1}{8}$